The equation of the plane which contains the y-axis and passes through the point (1, 2, 3) is :

Solution

Maths sir · 3 steps

A plane holding the y-axis has no y-term and no constant, so it is just ax + cz = 0.

1. Shape of the plane. The y-axis passes through the origin, so the plane does too, and the constant term is 0. The plane contains every point (0, t, 0), so the coefficient of y must also be 0.

   ax + cz = 0

2. Use the point. Put in (1, 2, 3): a + 3c = 0, so a = -3c. Taking c = -1 gives a = 3, and the plane is 3x - z = 0.

   Option	Through origin?	Value at (1, 2, 3)	Contains y-axis and the point?
   (a) x + 3z = 0	yes	1 + 9 = 10 ≠ 0	no
   (b) 3x - z = 0	yes	3 - 3 = 0	yes
   (c) x + 3z = 10	no	1 + 9 = 10	no, only parallel to the y-axis
   (d) 3x + z = 6	no	3 + 3 = 6	no, only parallel to the y-axis

3. So. Only 3x - z = 0 passes through both the origin and (1, 2, 3) with no y-term, so the answer is (b).

The idea to keep. Containing an axis means two conditions: drop that axis's variable and drop the constant term. A plane with no y-term but a nonzero constant is only parallel to the y-axis.

Answer from the NTA final answer key. Solution by PrepWiser.